- #1
Terry Bing
- 48
- 6
- Homework Statement
- The focal length of a spherical mirror is given by ##\frac{1}{f}=\frac{1}{v}+\frac{1}{u}## . If ##u=30\pm 0.3## cm and ##v=60\pm 0.6## cm. Find the percentage error in the measurement of focal length of the mirror .
- Relevant Equations
- We estimate errors by using formulae for differentials.
If I write ##f=\frac{uv}{u+v}## and then take differentials on both sides, I get ##\frac{df}{f}=\frac{du}{u}+\frac{dv}{v}+\frac{du+dv}{u+v}##, I get the fractional error as 0.03. (I have replaced the negative signs that come as a result of quotient rule with positive signs, since we are asked to estimate the maximum error. This leads to the formula that when measurements are multiplied or divided together, their relative errors just add up.)
If instead I write ##\frac{1}{f}=\frac{1}{v}+\frac{1}{u}## and then take differentials on both sides, then I get ##\frac{df}{f^2}=\frac{dv}{v^2}+\frac{du}{u^2}##. Using this, I get the fractional error to be 0.01. Which of these is a better estimate of the error? Also, why are there two different answers based on how I write the function. Basically I am taking first order approximations in both cases, right?
If instead I write ##\frac{1}{f}=\frac{1}{v}+\frac{1}{u}## and then take differentials on both sides, then I get ##\frac{df}{f^2}=\frac{dv}{v^2}+\frac{du}{u^2}##. Using this, I get the fractional error to be 0.01. Which of these is a better estimate of the error? Also, why are there two different answers based on how I write the function. Basically I am taking first order approximations in both cases, right?
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